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SIZIF [17.4K]
3 years ago
6

A compound in a copper ore has the following percentage composition by mass:

Chemistry
1 answer:
AveGali [126]3 years ago
6 0
1) since we are given percentages, we can assume we have 100 grams of the molecule.

55.6 % Cu ----> 55.6 grams Cu
16.4 % Fe------> 16.4 grams Fe
28.0% S--------> 28.0 grams S

2) convert each gram to moles using the molar masses given

55.6 g Cu \frac{1 mol}{63.5 g} = 0.876 mol Cu
16.4 g Fe \frac{1 mol}{56.0 g} = 0.293 mol Fe
28.0 gS \frac{1 mol}{32.0} = 0.875 mol S

3) we divide the smallest value of moles (0.293) to each one.

Cu --> 0.876 / 0.293= 3
Fe---> 0.293 / 0.293= 1
S-----> 0.875 / 0.293= 3

4) let's write the empirical formula

Cu₃FeS₃
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<h3><u>❄️</u></h3>

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4 years ago
If the sun rises at 7:36am and sets at 6:00pm, what times would we use for our data in the 24-hour clock format?
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6 0
3 years ago
Explain why materials with metallic lattice structures can be used to make wires and connections that conduct electricity in ele
kogti [31]
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7 0
3 years ago
1.72 moles of NOCI were placed in a 2.50 L reaction chamber
elena-14-01-66 [18.8K]

The equilibrium constant, Kc=0.026

<h3>Further explanation</h3>

Given

1.72 moles of NOCI

1.16 moles of NOCI  remained

2.50 L reaction chamber

Reaction

2NOCI(g) = 2NO(g) + Cl2(g).

Required

the equilibrium constant, Kc

Solution

ICE method

   2NOCI(g) = 2NO(g) + Cl2(g).

I    1.72

C   0.56           0.56         0.28

E   1.16              0.56         0.28

Molarity at equilibrium :

NOCl :

\tt \dfrac{1.16}{2.5}=0.464

NO :

\tt \dfrac{0.56}{2.5}=0.224

Cl2 :

\tt \dfrac{0.28}{2.5}=0.112

\tt Kc=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}\\\\Kc=\dfrac{0.224^2\times 0.112}{0.464^2}=0.026

4 0
3 years ago
Given the following balanced equation at 120°C: A(g) + B(g) ⇋ 2 C(g) + D(s)(a) At equilibrium a 4.0 liter container was found to
BlackZzzverrR [31]

Answer:

a) kc = 0,25

b) [A] = 0,41 M

c) [A] = <em>0,8 M</em>

[B] =<em>0,2 M</em>

[C] = <em>0,2M</em>

Explanation:

The equilibrium-constant expression is defined as the ratio of the concentration of products over concentration of reactants. Each concentration is raised to the power of their coefficient.

Also, pure solid and liquids are not included in the equilibrium-constant expression because they don't affect the concentration of chemicals in the equilibrium.

If global reaction is:

A(g) + B(g) ⇋ 2 C(g) + D(s)

The kc = \frac{[C]^2}{[A][B]}

a) The concentrations of each compound are:

[A] = \frac{1,60 moles}{4,0 L} = <em>0,4 M</em>

[B] = \frac{0,40 moles}{4,0 L} = <em>0,1 M</em>

[C] = \frac{0,40 moles}{4,0 L} = <em>0,1 M</em>

<em>kc = </em>\frac{[0,1]^2}{[0,4][0,1]} = 0,25

b) The addition of B and D in the same amount will, in equilibrium, produce these changes:

[A] = \frac{1,60-x moles}{4,0 L}

[B] = \frac{0,60-x moles}{4,0 L}

[C] = \frac{0,60+2x moles}{4,0 L}

0,25 = \frac{[0,60+2x]^2}{[1,60-x][0,60-x]}

You will obtain

3,75x² +2,95x +0,12 = 0

Solving

x =-0,74363479081119   → No physical sense

x =-0,043031875855476

Thus, concentration of A is:

\frac{1,60-(-0,04 moles)}{4,0 L} = <em>0,41 M</em>

c) When volume is suddenly halved concentrations will be the concentrations in equilibrium over 2L:

[A] = \frac{1,60 moles}{2,0 L} = <em>0,8 M</em>

[B] = \frac{0,40 moles}{2,0 L} = <em>0,2 M</em>

[C] = \frac{0,40 moles}{2,0 L} = <em>0,2M</em>

I hope it helps!

8 0
4 years ago
Read 2 more answers
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